Find the optimal solution for the following problem. (Round your answers to 3 decimal places.)

Mathematics

1 answer:

Sphinxa [80]2 years ago

3 0

Answer:

x=2.125

y=0

C=19.125

Step-by-step explanation:

To solve this problem we can use a graphical method, we start first noticing the restrictions $x\geq 0$ and $y\geq 0$ , which restricts the solution to be in the positive quadrant. Then we plot the first restriction $8x+10y\leq 17$ shown in purple, then we can plot the second one $11x+12y\leq 25$ shown in the second plot in green.

The intersection of all three restrictions is plotted in white on the third plot. The intersection points are also marked.

So restrictions intersect on (0,0), (0,1.7) and (2.215,0). Replacing these coordinates on the objective function we get C=0, C=11.9, and C=19.125 respectively. So The function is maximized at (2.215,0) with C=19.125.

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Answer

a) C

b) B

c) C, E

d) 15.5

e) 0.7

f) 67.7

Step-by-step explanation

a) A linear equation can be written as $y=mx+c$ where $m$ is the slope and $c$ is the intercept on the $y$ -axis. The slope describes how $y$ changes when $x$ changes. A negative value means the it is a decreasing change. In our case, $y$ is $A$ and $x$ is $t$ . Thus, the slope of the equation is $m=-0.062$ . This means that the area of glacier is decreasing by $0.062 \text{km}{}^2 = 62000 \text{m}{}^2$ every year since 2000.

b) The $A$ -intercept (= 16.2 $\text{km}{}^2$ ) represents the area covered when $t=0$ . But $t=0$ starting from year 2000. This intercept then represents the area covered by glacier in the year 2000. Therefore, the area covered by glacier in 2000 was 16.2 $\text{km}{}^2$ .

c) $A=f(t)$ means the area covered after year 2000. Setting it to 12 means the area covered after $t$ years is 12 $\text{km}{}^2$ ). $t$ is therefore the number of years since 2000 that the total area covered will be 12 $\text{km}{}^2$ ).

d) $f(12)=16.2−0.062\times12 =16.2-0.744 = =15.456 15.5 \text{km}{}^2$ to 1 decimal place.

e) The term involving $t$ represents the disappearing area. In 12 years, the disappeared area is $0.062\times12=0.744 =0.7 \text{km}{}^2$ to 1 decimal place.